Answer :
Answer:
0.472 grams of aluminum oxide would be produced from the reaction of 0.25 grams of aluminum with excess oxygen gas.
Explanation:
First of all, you should know that the balanced reaction that occurs:
4 Al + 3 O₂ → 2 Al₂O₃
Then, by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts are produced in moles:
- Al: 4 moles
- O₂: 3 moles
- Al₂O₃: 2 moles
Being:
- Al: 27 g/mole
- O: 16 g/mole
The molar mass of the compounds participating in the reaction is:
- Al: 27 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- Al₂O₃: 2*27 g/mole + 3*16 g/mole= 102 g/mole
Then, by stoichiometry of the reaction, the following amounts of mass react and are produced:
- Al: 4 moles*27 g/mole= 108 g
- O₂: 3 moles*32 g/mole= 96 g
- Al₂O₃: 2 moles*102 g/mole= 204 g
Now it is possible to apply the following rule of three: if 108 grams of Al produce 204 grams of Al₂O₃, 0.25 grams of aluminum how much mass of Al₂O₃ will it produce?
[tex]mass of Al_{2} O_{3} =\frac{0.25 grams of Al*204 grams of Al_{2} O_{3} }{108 grams of Al}[/tex]
mass of Al₂O₃=0.472 grams
0.472 grams of aluminum oxide would be produced from the reaction of 0.25 grams of aluminum with excess oxygen gas.